The distance of the point P(1, 2, 1) from the | vedclass

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(C) The line along which the distance is measured is parallel to the line $\frac{x - 5}{1} = \frac{y - 3/2}{1} = \frac{z - 5/2}{1}$.Thus,the direction

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$\frac{7\sqrt{3}}{2}$

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(C) The line along which the distance is measured is parallel to the line $\frac{x - 5}{1} = \frac{y - 3/2}{1} = \frac{z - 5/2}{1}$.Thus,the direction ratios of the line are $(1, 1, 1)$.The equation of the line passing through $P(1, 2, 1)$ with direction ratios $(1, 1, 1)$ is:$\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 1}{1} = \lambda$Any point $Q$ on this line is given by $Q(\lambda + 1, \lambda + 2, \lambda + 1)$.Since $Q$ lies on the plane $2x + y - z = 10$,we have:$2(\lambda + 1) + (\lambda + 2) - (\lambda + 1) = 10$$2\lambda + 2 + \lambda + 2 - \lambda - 1 = 10$$2\lambda + 3 = 10$$2\lambda = 7 \Rightarrow \lambda = \frac{7}{2}$Substituting $\lambda = \frac{7}{2}$ in the coordinates of $Q$,we get $Q\left(\frac{9}{2}, \frac{11}{2}, \frac{9}{2}ight)$.The distance $PQ$ is the distance between $P(1, 2, 1)$ and $Q\left(\frac{9}{2}, \frac{11}{2}, \frac{9}{2}ight)$:$PQ = \sqrt{\left(\frac{9}{2} - 1ight)^2 + \left(\frac{11}{2} - 2ight)^2 + \left(\frac{9}{2} - 1ight)^2}$$PQ = \sqrt{\left(\frac{7}{2}ight)^2 + \left(\frac{7}{2}ight)^2 + \left(\frac{7}{2}ight)^2}$$PQ = \sqrt{3 	imes \frac{49}{4}} = \frac{7\sqrt{3}}{2}$

The distance of the point $P(1, 2, 1)$ from the plane $2x + y - z = 10$ measured along the line $\frac{x - 5}{1} = \frac{2y - 3}{2} = \frac{z - \frac{5}{2}}{1}$ is

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